When perpendicular becomes concurrent

The Asian Pacific Mathematical Olympiad (APMO) is a long-running regional olympiad that challenges young mathletes across the region with a handful of hard, original problems each year.[^1] Problem 4 from this year's edition is a nice showcase of a common trick: restating an unfriendly definition into a more symmetric one turns the perpendicularity we want into a friendlier concurrence, and the solution follows naturally.

[^1]: For more information and the rest of this year's problems, go to apmo-official.org.

Problem 4. Let $ABCD$ be a quadrilateral with an incircle $\omega$ of center $I$. The diagonals $AC$ and $BD$ intersect at $E$. Let $J$ be the incenter of triangle $ABD$. The extension of the ray $EJ$ intersects $\omega$ at $P$. Prove that $PI \perp BD$.

{: w = "100"}

Solution. First, let's modify the statement. It seems more intuitive to let $P$ be the endpoint of the diameter of $\omega$ perpendicular to $BD$ that lies in the interior of $\bigtriangleup ABD$. Thus, the problem reduces to showing that the points $P,\ J$, and $E$ are collinear.

Another key point. The triangle $\bigtriangleup ABD$ appears to be an arbitrary choice. In other words, the statement should remain valid for $\bigtriangleup BCD$ too. In this spirit, define $F$ and $P'$ to be the incenter of $\bigtriangleup BCD$ and the intersection of $PI$ with $\omega$ that lies within $\bigtriangleup BCD$ (i.e., the point diametrically opposite to $P$ in $\omega$), respectively. Then, it should also be true that $P', F$, and $E$ are collinear. With these preparations, we're ready to tackle the problem at hand. Consider the following

Claim 1. The line $JF$ is perpendicular to $BD$.

Proof. Let $R$ and $R'$ be the points where the incircles of $\bigtriangleup ABD$ and $\bigtriangleup BCD$ touch $BD$. It's known that

$DR = s_{ABD}-AB = \frac{DA+BD-AB}{2},\quad DR' = s_{BCD} - BC= \frac{CD+BD-BC}{2}$

where $s_{XYZ}$ stands for the semiperimeter of $\bigtriangleup XYZ$. Pitot's theorem implies straightforwardly that $\frac{DA+BD-AB}{2}=\frac{CD+BD-BC}{2}$. Hence, $DR=DR'$, so $R\equiv R'$ and thus $J,\ R$, and $F$ are collinear with $JF\perp BD$. $\square$

{: w = "100"}

With this result, we have already used the fact that $ABCD$ has an incircle and the definition of $J$. We need to exploit the redefinition of $P$. Let's now argue that

Claim 2. The lines $AP$, $CP'$ and $BD$ concur.

Proof. Let $R'$ be the point diametrically opposite to $R$ in the incircle of $\bigtriangleup ABD$. As $A$ is the exsimilicenter of this circle and $\omega$, it turns out that $R'$ lies on $AP$. More importantly, it's known that the point $T\equiv \overline{AR'}\cap \overline{BD}$ is the contact point of the $A$-excircle of $\bigtriangleup ABD$ with $BD$ and that $BT=DR$. Thus, this point $T$ lies on $AP$ and satisfies that $BT=DR$. Analogously, we can prove that $CP'$ meets $BD$ at a point $T'$ where $BT'=DR$, which forces $T\equiv T'$, implying the desired concurrence. $\square$

We're ready to conclude. By Claim 1, observe that $PP'\parallel JF$. Together with Claim 2, we find that:

$\frac{PS}{SP'} = \frac{R'R}{RR''} = \frac{JR}{RF}$

with $S\equiv \overline{PP'}\cap\overline{BD}$ and $R''$ the antipode of $R$ in the incircle of $\bigtriangleup BCD$. This result and Thales' theorem jointly imply that $PJ$ and $P'F$ meet each other at a point on $BD$, say $E'$. Let $M$ be the midpoint of the segment $JF$. Of course, Thales' theorem ensures that $IM$ passes through $E'$. Now, it is also known that the pole of $BD$ with respect to $\omega$ is $Q\equiv \overline{AC}\cap \overline{PI}$ and that $A,C,E,Q$ forms a harmonic bundle. Let $M'\equiv \overline{JF}\cap \overline{IE}$. Then:

$-1=(A,C;E,Q) \overset{I}{=} (J,F;M',P)$

which together with $JF\parallel PP'$ implies that $M'$ is the midpoint of segment $JF$, i.e., $M\equiv M'$. Hence $E\equiv E'$ and the required collinearity follows. $\square$